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frozen [14]
3 years ago
13

A gas occupies 4.00 L at 2.00 atm pressure. At what pressure would the volume be 6.00 L if the temperature remains constant?

Chemistry
2 answers:
kherson [118]3 years ago
8 0
P1V1= P2V2 
P1= 2
V1=4
V2=6
answere= (8/6)
True [87]3 years ago
6 0

Answer:

The final pressure would be 1.33 atm

Explanation:

<u>Given:</u>

Initial volume of the gas, V1 = 4.00 L

Initial pressure, P1 = 2.00 atm

Final volume, V2 = 6.00 L

<u>To determine:</u>

The final pressure P2 of the gas

<u>Explanation:</u>

Based on the ideal gas equation:

PV = nRT

where: n = number of moles,

R = gas constant and T = temperature

At constant Temperature the gas law reduces to,

PV = constant\\\\P1V1 = P2V2\\\\P2 = \frac{P1V1}{V2} = \frac{2.00atm*4.00L}{6.00L} =1.33\ atm

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Answer:

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In order to prepare 10 mL, 15 μM; <em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM; <em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

Explanation:

Using the dilution equation:

no of moles before dilution = no of moles after dilution.

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25 x initial volume = 5 x 10

Initial volume = 50/25

                       = 2 mL

<em>2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

<em />

In order to prepare 10 mL, 10 μM from 25 μM stock,

25 x initial volume = 10 x 10

Initial volume = 100/25 = 4 mL

<em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

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25 x initial volume = 15 x 10

initial volume = 150/25 = 6 mL

<em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

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<em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

6 0
3 years ago
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luda_lava [24]
Answer to this question is C. Regarding the volume.
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Answer:

A

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7 0
1 year ago
Read 2 more answers
A. 20 moles of NH3 are needed to produce how many moles of H2O?
tiny-mole [99]

a. 30 moles of H₂O

b. 2.33 moles of N₂

<h3>Further explanation</h3>

Given

a. 20 moles of NH₃

b. 3.5 moles of O₂

Required

a. moles of H₂O

b. moles of N₂

Solution

Reaction

4NH₃+3O₂⇒2N₂+6H₂O

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=6/4 x mol NH₃

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b. From the equation, mol ratio N₂ : O₂ = 2 : 3, so mol N₂ :

=2/3 x mol O₂

= 2/3 x 3.5 moles

= 2.33 moles

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2 years ago
Explain how are elements classified and provide an example
swat32
By nonmetals, metals, and gases
4 0
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